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In some platform, int32_t (from stdint.h) is long int, but in other platform, it could be int. When I want to use printf, how can I determine which format, "%ld" or "%d", should be used?
Or, perhaps, I should force converting it to long like below:
int32_t m;
m = 3;
printf ("%ld\n", (long)m);
But that solution is tedious. Any suggestions?
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In C (since C99), the inttypes. h contains macros that expand to format specifiers for the fixed-width types. For int32_t : printf("%" PRId32 "\n", m);
%d is a signed integer, while %u is an unsigned integer. Pointers (when treated as numbers) are usually non-negative. If you actually want to display a pointer, use the %p format specifier.
In C programming language, %d and %i are format specifiers as where %d specifies the type of variable as decimal and %i specifies the type as integer. In usage terms, there is no difference in printf() function output while printing a number using %d or %i but using scanf the difference occurs.
The Printf module API details the type conversion flags, among them: %B: convert a boolean argument to the string true or false %b: convert a boolean argument (deprecated; do not use in new programs).
In C (since C99), the inttypes.h contains macros that expand to format specifiers for the fixed-width types. For int32_t:
printf("%" PRId32 "\n", m);
That macro is likely to expand to "d" or "ld". You can put the usual modifiers and so on, e.g.:
printf("%03" PRId32 "\n", m);
In C++ (since C++11) the same facility is available with #include <inttypes.h>, or #include <cinttypes>.
Apparently, some C++ implementations require the user to write #define __STDC_FORMAT_MACROS 1 before #include <inttypes.h>, even though the C++ Standard specifies that is not required.
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